### 2022 USEMO #3

Point $P$ lies in the interior of a triangle $ABC$. Lines $AP$, $BP$, and $CP$ meet the opposite sides of triangle $ABC$ at $A$', $B'$, and $C'$ respectively. Let $P_A$ the midpoint of the segment joining the incenters of triangles $BPC'$ and $CPB'$, and define points $P_B$ and $P_C$ analogously. Show that if $AB'+BC'+CA'=AC'+BA'+CB'$then points $P,P_A,P_B,$ and $P_C$ are concyclic.

Let $X_A$, $X_B$, $X_C$ be the projections of $P_A$, $P_B$, $P_C$ onto $\overline{BB'}$, $\overline{CC'}$, $\overline{AA'}$, respectively. Then$PX_A=\tfrac14(BP+PC'-C'B)-\tfrac14(B'P+PC-CB'),$so (with lengths appropriately directed) we have $PX_A+PX_B+PX_C=0$. Let $\alpha=\angle BPP_A$, and similarly define $\beta$ and $\gamma$, so $\alpha+\beta+\gamma=90^\circ$. Let $P_A'$, $P_B'$, $P_C'$ be the inverses of $P_A$, $P_B$, $P_C$ about $P$ with radius 1. Then\begin{align*}     \operatorname{Area}(\triangle P_A'P_B'P_C') &=\sum_\mathrm{cyc}\operatorname{Area}(\triangle PP_B'P_C') =\sum_\mathrm{cyc} PP_B'\cdot PP_C'\cdot\sin(\alpha+\beta)\\ &=\sum_\mathrm{cyc}\frac{\cos\alpha}{PX_B}\cdot\frac{\cos\beta}{PX_C}\cdot\cos\gamma =\frac{\cos\alpha\cos\beta\cos\gamma}{PX_A\cdot PX_B\cdot PX_C}\cdot\sum_\mathrm{cyc} PX_A=0. \end{align*}It follows that $P_A'$, $P_B'$, $P_C'$ are collinear, as desired. $\square$